Responding to a pm:

You need to understand the concepts of mean, median and range very well to do this question. Since you asked me to resolve it, I am assuming that you are comfortable with these stats concepts I have discussed on my blog. Here is the logical solution:

"the difference between the highest and lowest salaries is $100,000."
So there are at least 2 people - say one with salary 0 and the other with 100k. No salary will be outside this range.

Median = 50k more than lowest. So median is right in the center of lowest and highest since lowest and highest differ by 100k. In our example, median = 50k. Since there are more than 2 people, there would probably be a person at 50k.

Mean = 20k more than median so in our example, mean salary = 70k

On the number line,

0........50k (median)........100k

Mean = 70k

So there must be people more toward 100k to bring the mean up to 70k. Since we want to add minimum people, we will add people at 100k to quickly make up the right side deficit. 0 and 50k are (70k + 20k) = 90k away from 70k. 100k is 30k away from 70k. To bring the mean to 70k, we will add two people at 100k each to get:

0....50k.....100k, 100k, 100k

But when we add more people to the right of 70k, the median will shift to the right. We need to keep the median at 50k. So every time we add people to the right of 70k, we need to add people at 50k too to balance the median. 50k is 20k less than 70k while 100k is 30k more than 70k. To keep the mean same, we need to add 2 people at 100k for every 3 people we add at 50k. So if we add 3 people at 50k and 2 people at 100k, we get:

0, ... 50k, 50k, 50k, 50k, ... 100k, 100k, 100k, 100k, 100k
the median is not at 50k yet.

Add another 3 people at 50k and another 2 at 100k to get

0, 50k, 50k, 50k, 50k, 50k, 50k, 50k, 100k, 100k, 100k, 100k, 100k, 100k, 100k

Now the median is 50k and mean is 70k.

Total number of people is 15.

Answer (C)
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Hello Karishma,

That is a great solution. Unfortunately, I am unable to understand the concept behind this. Kindly give the link of your blog as reference. That would be beneficial for me.

We can also use POE .Median is the middle number between lowest and highest so total number of students will be odd .This eliminates choices A,B,D
Between C and E .Average is 70000 which is more towards higher pay and is also more then median .So number of students will be more on the rhs then lhs .Add the students accordingly on both sides to get the final answer.

Regards,
Manish Khare
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Let the lowest salary be x. Then, the median salary is x + 50,000; the average salary is x + 70,000 and the greatest salary is x + 100,000.

We notice that the average salary is between the median and the greatest salary. When this is the case, the number of students in the class can be minimized by choosing the salaries as one salary which is equal to x, some number of salaries which are equal to x + 50,000 and some number of salaries which are equal to x + 100,000. The reason for such a choice of salaries is because every salary we add that is less than x + 50,000 will bring the average salary down, and to counter that, since we can’t add any salaries greater than x + 100,000, we have to increase the number of salaries that equal x + 100,000, therefore increasing the number of students.

So, let n be the number of salaries that equal x + 50,000 and m be the number of salaries that equal x + 100,000. Notice that m cannot be greater than n because otherwise, the median will equal x + 100,000. Since the average salary is x + 70,000 and the number of students is n + m + 1, the sum of all the salaries is (n + m + 1)(x + 70,000). On the other hand, the sum of all the salaries is also equal to x + n(x + 50,000) + m(x + 100,000). Let’s equate these two quantities:

x + n(x + 50,000) + m(x + 100,000) = (n + m + 1)(x + 70,000)

x + nx + 50,000n + mx + 100,000m = nx + mx + x + 70,000n + 70,000m + 70,000

30,000m = 20,000n + 70,000

3m = 2n + 7

Since m ≤ n, 3m ≤ 3n. Let’s substitute 3m = 2n + 7:

2n + 7 ≤ 3n

7 ≤ n

We see that the smallest value of n is 7. If n = 7, then 3m = 14 + 7 = 21; therefore m = 7 as well. Thus, the smallest number of students is n + m + 1 = 7 + 7 + 1 = 15.

Answer: C
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